# Round Transformations

This page gives a description of the four operations that compose a round of AES. Each has been designed to satisfy criterias, one of them is that all must be invertible.

This is the only operation that involves a key. It is obvious that omitting it would mean no encryption.

Round keys are derived from the master key (see the Key Schedule section) and are all composed of 16 bytes. We simply xor byte by byte the state by the bytes of the round key in the according position.

Its inverse is itself: if we xor again, we get back the original value of the state.

# Mix Columns

A major goal of MC is the diffusion property: each byte has an impact on the whole column, so a modification propagates to other positions.

This operation mixes each column of the state independently from each other: each byte of the column is replaced by a (slightly different) combination of the four bytes of the column.

Let $(a_0, a_1, a_2, a_3)$ the quadruplet of elements of a column, the operation MC is done by multiplying with a matrix $M$.

The calculations are performed in the finite field. If you are not familiar enough with the notions, you can skip to the next part and retain that this operation is also invertible using another matrix.

This matrix is circulant: each row is the same as the one above but is shifted by one column. So we can construct it in one line of SageMath (recall that the bytes 02 and 03 correspond respectively to $x$and $x+1$ in the field):

# we construct the matrix# (we use the finite field F constructed in the previous page)M = Matrix.circulant([x, x + 1, 1, 1])M# [    x x + 1     1     1]# [    1     x x + 1     1]# [    1     1     x x + 1]# [x + 1     1     1     x]​# We multiply a columncol1 = vector([F.random_element() for i in range(4)])col1# (x^6 + x + 1, x^7 + x^4 + x^2, x^6 + x^4 + x^3 + x^2, x^7 + x^6 + x^4 + x^3 + x^2)col2 = M*col1col2# (x^7 + x^5 + 1, x^6 + x^3, x^4, x^7 + x^5 + x^3 + x^2 + x)

## Inverse of Mix Column

It is needed to reverse MC for the decryption process. Fortunately, there exists a matrix $N$ such that $M\times N = N \times M = \textrm{Id}$, the identity matrix.

N = M^-1N# [x^3 + x^2 + x   x^3 + x + 1 x^3 + x^2 + 1       x^3 + 1]# [      x^3 + 1 x^3 + x^2 + x   x^3 + x + 1 x^3 + x^2 + 1]# [x^3 + x^2 + 1       x^3 + 1 x^3 + x^2 + x   x^3 + x + 1]# [  x^3 + x + 1 x^3 + x^2 + 1       x^3 + 1 x^3 + x^2 + x]​# hexadecimal representationfor row in N:    print([F_to_hex(a) for a in row])# ['e', 'b', 'd', '9']# ['9', 'e', 'b', 'd']# ['d', '9', 'e', 'b']# ['b', 'd', '9', 'e']​col3 = N*col2col3 == col1# True

We remark that the coefficient of the matrix are less friendly for the inverse operation as it involves polynomial of higher degree. This means that the decryption is a bit slower.

# Shift Rows

The goal of shifting rows is reinforce the diffusion property: the bytes of a column are shifted so they are all positioned on different columns. Combined with MC, then a one byte modification will have an effect to the whole state after several rounds: this is the Avalanche effect.

This operation shifts the rows of the state in the following manner:

The rows are shifted from top to bottom respectively by an offset of 0, 1, 2 or 3 columns to the left. Bytes that overflow on the left are put back to the right of the row.

The inverse is almost the same: the rows are shifted to the right instead by the same offsets and exceeding bytes are put back to the left.

Small exercise: to what extent would be the impact on the security of AES if no shifting were present?

# Substitution Box

Last, but not least, the SB design criterias is to bring the confusion property to make correlation between a plaintext and a ciphertext as small as possible.

The precedent operations shuffle the plaintext in such a way that any modification of a byte has an impact to the whole state after several rounds. Though, this is not enough as those operations are linear: it means that the ciphertext could be expressed as linear equations from the plaintext and the master key. From a known couple plaintext/ciphertext it would be easy to solve the system and find the key.

The substitution box is the operation that breaks the linearity: each byte of the state is replaced by another following the table below.

 ​ 00 01 02 03 04 05 06 07 08 09 0a 0b 0c 0d 0e 0f 00 63 7c 77 7b f2 6b 6f c5 30 01 67 2b fe d7 ab 76 10 ca 82 c9 7d fa 59 47 f0 ad d4 a2 af 9c a4 72 c0 20 b7 fd 93 26 36 3f f7 cc 34 a5 e5 f1 71 d8 31 15 30 04 c7 23 c3 18 96 05 9a 07 12 80 e2 eb 27 b2 75 40 09 83 2c 1a 1b 6e 5a a0 52 3b d6 b3 29 e3 2f 84 50 53 d1 00 ed 20 fc b1 5b 6a cb be 39 4a 4c 58 cf 60 d0 ef aa fb 43 4d 33 85 45 f9 02 7f 50 3c 9f a8 70 51 a3 40 8f 92 9d 38 f5 bc b6 da 21 10 ff f3 d2 80 cd 0c 13 ec 5f 97 44 17 c4 a7 7e 3d 64 5d 19 73 90 60 81 4f dc 22 2a 90 88 46 ee b8 14 de 5e 0b db a0 e0 32 3a 0a 49 06 24 5c c2 d3 ac 62 91 95 e4 79 b0 e7 c8 37 6d 8d d5 4e a9 6c 56 f4 ea 65 7a ae 08 c0 ba 78 25 2e 1c a6 b4 c6 e8 dd 74 1f 4b bd 8b 8a d0 70 3e b5 66 48 03 f6 0e 61 35 57 b9 86 c1 1d 9e e0 e1 f8 98 11 69 d9 8e 94 9b 1e 87 e9 ce 55 28 df f0 8c a1 89 0d bf e6 42 68 41 99 2d 0f b0 54 bb 16

If 3b is the input, its output is on row 30 and column 0b and is the byte e2.

Let sbox the name of this table, then for any bytes b1 and b2, we have

$\texttt{sbox}(\texttt{b1} \oplus \texttt{b2}) \neq \texttt{sbox}(\texttt{b1}) \oplus \texttt{sbox}(\texttt{b2})$

which is the desired property.

Though, this is not enough and the design of the sbox is made to include a sufficient algebraic complexity. The construction of the sbox table is done in two steps:

1. An element$a$is replaced by $b = a^{-1}$ ($0$ has no inverse and is mapped to $0$);

2. An affine transformation on the coefficients of $a$:

$\begin{bmatrix} 1 & 0 & 0 & 0 & 1 & 1 & 1 & 1 \\ 1 & 1 & 0 & 0 & 0 & 1 & 1 & 1 \\ 1 & 1 & 1 & 0 & 0 & 0 & 1 & 1 \\ 1 & 1 & 1 & 1 & 0 & 0 & 0 & 1 \\ 1 & 1 & 1 & 1 & 1 & 0 & 0 & 0 \\ 0 & 1 & 1 & 1 & 1 & 1 & 0 & 0 \\ 0 & 0 & 1 & 1 & 1 & 1 & 1 & 0 \\ 0 & 0 & 0 & 1 & 1 & 1 & 1 & 1 \end{bmatrix} \cdot \begin{bmatrix} b_0 \\ b_1 \\ b_2 \\ b_3 \\ b_4 \\ b_5 \\ b_6 \\ b_7 \end{bmatrix} + \begin{bmatrix} 1 \\ 1 \\ 0 \\ 0 \\ 0 \\ 1 \\ 1 \\ 0 \end{bmatrix}$

The table we gave above has been construted with SageMath and applying the two steps:

def construct_sbox():    mat = Matrix.circulant(vector(GF(2), [1,0,0,0,1,1,1,1]))    sbox = []    for a in range(256):        # convert a byte value to field element        b = sum(((a >> i) & 1)*x^i for i in range(8))        # first step: map to inverse        if b != 0:        b = b^-1        # second step:        # affine transformation using a vector space over GF(2)        b = mat*vector(b) + vector(GF(2), [1, 1, 0, 0, 0, 1, 1, 0])        # convert the vector as an integer in [0, 255]        sbox.append(sum(ZZ(b[i]) << i for i in range(8)))    return sbox​sbox = construct_sbox()for i in range(16):    print([f'{a:02x}' for a in sbox[16*i : 16*(i + 1)]])# ['63', '7c', '77', '7b', 'f2', '6b', '6f', 'c5', '30', '01', '67', '2b', 'fe', 'd7', 'ab', '76']# ['ca', '82', 'c9', '7d', 'fa', '59', '47', 'f0', 'ad', 'd4', 'a2', 'af', '9c', 'a4', '72', 'c0']# ['b7', 'fd', '93', '26', '36', '3f', 'f7', 'cc', '34', 'a5', 'e5', 'f1', '71', 'd8', '31', '15']# ['04', 'c7', '23', 'c3', '18', '96', '05', '9a', '07', '12', '80', 'e2', 'eb', '27', 'b2', '75']# ['09', '83', '2c', '1a', '1b', '6e', '5a', 'a0', '52', '3b', 'd6', 'b3', '29', 'e3', '2f', '84']# ['53', 'd1', '00', 'ed', '20', 'fc', 'b1', '5b', '6a', 'cb', 'be', '39', '4a', '4c', '58', 'cf']# ['d0', 'ef', 'aa', 'fb', '43', '4d', '33', '85', '45', 'f9', '02', '7f', '50', '3c', '9f', 'a8']# ['51', 'a3', '40', '8f', '92', '9d', '38', 'f5', 'bc', 'b6', 'da', '21', '10', 'ff', 'f3', 'd2']# ['cd', '0c', '13', 'ec', '5f', '97', '44', '17', 'c4', 'a7', '7e', '3d', '64', '5d', '19', '73']# ['60', '81', '4f', 'dc', '22', '2a', '90', '88', '46', 'ee', 'b8', '14', 'de', '5e', '0b', 'db']# ['e0', '32', '3a', '0a', '49', '06', '24', '5c', 'c2', 'd3', 'ac', '62', '91', '95', 'e4', '79']# ['e7', 'c8', '37', '6d', '8d', 'd5', '4e', 'a9', '6c', '56', 'f4', 'ea', '65', '7a', 'ae', '08']# ['ba', '78', '25', '2e', '1c', 'a6', 'b4', 'c6', 'e8', 'dd', '74', '1f', '4b', 'bd', '8b', '8a']# ['70', '3e', 'b5', '66', '48', '03', 'f6', '0e', '61', '35', '57', 'b9', '86', 'c1', '1d', '9e']# ['e1', 'f8', '98', '11', '69', 'd9', '8e', '94', '9b', '1e', '87', 'e9', 'ce', '55', '28', 'df']# ['8c', 'a1', '89', '0d', 'bf', 'e6', '42', '68', '41', '99', '2d', '0f', 'b0', '54', 'bb', '16']

The inverse of the table is simple to produce: we only need to reverse the match between an input and its output.

Exercise: write the inverse of the substitution box in SageMath using the inverse of the affine transformation.